Will you square? Part 3

Algebra Level 3

If $\sqrt { 2 } +\sqrt { 5 }$ is written in the form $\frac { a }{ \sqrt { b-\sqrt { c } } },$ where $a,b,c$ are positive integers and $b$ is not divisible by the square of any prime, what is the value of $a+b+c?$

The answer is 50.

4 solutions

Shivamani Patil
Oct 4, 2014

We have

$\sqrt { 2 } +\sqrt { 5 } =\sqrt { { \left( \sqrt { 2 } +\sqrt { 5 } \right) }^{ 2 } }$

$=\sqrt { 2+5+2\sqrt { 10 } }$

$=\sqrt { 7+\sqrt { 40 } }$

$=\sqrt { \frac { \left( 7+\sqrt { 40 } \right) \left( 7-\sqrt { 40 } \right) }{ \left( 7-\sqrt { 40 } \right) } }$

$=\sqrt { \frac { { 7 }^{ 2 }-{ (\sqrt { 40 } ) }^{ 2 } }{ \left( 7-\sqrt { 40 } \right) } }$

$=\sqrt { \frac { 9 }{ \left( 7-\sqrt { 40 } \right) } } =\frac { 3 }{ \sqrt { 7-\sqrt { 40 } } }$

$\Rightarrow a=3,b=7,c=40 \Rightarrow a+b+c=50$

$40 = 2^2 \times 10$ ? How is 40 square free? Did you mean c is not a square?

Siddhartha Srivastava - 6 years, 8 months ago

a number having two integral square roots is a perfect square

aaron paul - 6 years, 8 months ago

The question was changed after I replied. Hate when that happens. Before the change, the question read "... c is not squarefree ... ".

Siddhartha Srivastava - 6 years, 8 months ago

@Siddhartha Srivastava – Sorry for that.And thank u for pointing it.

shivamani patil - 6 years, 8 months ago

do not forget that a square has two roots so -3/x is also possible and the other possible answer is 44
(x stands for denominator)

aaron paul - 6 years, 8 months ago

He already stated that a, b, c are all positive integers.

Jared Low - 6 years, 5 months ago

Can't $(a,b,c)=(6, 28, 640)$ also be solution according to your conditions?

Jared Low - 6 years, 5 months ago

Thanks. I've updated the phrasing to "b is not divisible by the square of any prime".

Calvin Lin Staff - 6 years, 5 months ago

but if we just multiply and divide by conjugate we will get a different answer ie 44

Riddhesh Deshmukh - 5 years, 8 months ago

Chew-Seong Cheong
Oct 6, 2014

It is given that: $\dfrac {a}{\sqrt{b - \sqrt{c} }} = \sqrt{2} + \sqrt{5}$

Squaring both sides, we have:

$\dfrac {a^2}{b - \sqrt{c} } = 2 + 2\sqrt{2}\sqrt{5} + 5 = 7 + \sqrt{40}$

$\quad \quad = \dfrac {(7 + \sqrt{40} ) (7 - \sqrt{40} ) } {7 - \sqrt{40} } = \dfrac {9} {7 - \sqrt{40} }$

Square-rooting both sides, we have:

$\dfrac {a}{\sqrt{b - \sqrt{c} }} = \dfrac {3} {\sqrt{7 - \sqrt{40}} }$

$\Rightarrow a =3, b = 7, c = 40\quad \Rightarrow a+b+c = 3+7+40 = \boxed{50}$

Asad Jawaid
Jan 28, 2017

Naman Kapoor
Mar 26, 2015

We can just multiply and divide by its conjugate

Will you square? Part 3

The answer is 50.

4 solutions

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